| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | To solve this problem, we need to find the rate of change of Julian's y-coordinate (dy/dt) when his x-coordinate is changing at a rate of -1.25 m/s at the point (40, 30). First, we need to find the equation of the circular track. The equation of a circle with radius r and center at the origin (0, 0) is given by: x^2 + y^2 = r^2 Substituting the given radius of 50 m, we get: x^2 + y^2 = 2500 We are given that Julian's x-coordinate is changing at a rate of -1.25 m/s at the point (40, 30). We can find the rate of change of Julian's y-coordinate (dy/dt) by using implicit differentiation. Differentiating both sides of the equation with respect to time t, we get: 2x(dx/dt) + 2y(dy/dt) = 0 Substituting the given values of x, y, and dx/dt, we get: 2(40)(-1.25) + 2(30)(dy/dt) = 0 Simplifying, we get: -100 + 60(dy/dt) = 0 Solving for dy/dt, we get: dy/dt = 100/60 = 5/3 Therefore, the rate of change of Julian's y-coordinate (dy/dt) at the point (40, 30) when his x-coordinate is changing at a rate of -1.25 m/s is 5/3 m/s or approximately 1.67 m/s. |
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| deepseekr1 | Julian is jogging around a circular track with a radius of 50 meters. The equation of the circle is \(x^2 + y^2 = 2500\). To find the rate at which his y-coordinate is changing (\(dy/dt\)) when his coordinates are (40, 30) and his x-coordinate is changing at a rate of \(-1.25\) m/s, we start by differentiating the equation of the circle with respect to time \(t\): \[ \frac{d}{dt}(x^2 + y^2) = \frac{d}{dt}(2500) \] This gives: \[ 2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0 \] Dividing through by 2, we get: \[ x \frac{dx}{dt} + y \frac{dy}{dt} = 0 \] Substituting the given values \(x = 40\), \(y = 30\), and \(\frac{dx}{dt} = -1.25\): \[ 40(-1.25) + 30 \frac{dy}{dt} = 0 \] Solving for \(\frac{dy}{dt}\): \[ -50 + 30 \frac{dy}{dt} = 0 \] \[ 30 \frac{dy}{dt} = 50 \] \[ \frac{dy}{dt} = \frac{50}{30} = \frac{5}{3} \] Thus, the rate at which Julian's y-coordinate is changing at that moment is \(\boxed{\dfrac{5}{3}}\) m/s. |
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| gpto1 | **Solution:** We are given: - Julian is jogging around a circular track of radius \( r = 50 \) meters. - His position at a certain moment is \( (x, y) = (40, 30) \). - At that moment, his \( x \)-coordinate is changing at a rate of \( \dfrac{dx}{dt} = -1.25 \) m/s. - We are asked to find \( \dfrac{dy}{dt} \) at that moment. **Step 1: Write the Equation of the Circle** Since Julian is jogging around a circular track of radius 50 m centered at the origin, the equation of the circle is: \[ x^2 + y^2 = r^2 \] \[ x^2 + y^2 = 50^2 \] \[ x^2 + y^2 = 2500 \] **Step 2: Differentiate Both Sides with Respect to Time \( t \)** We will use implicit differentiation with respect to time \( t \): \[ \frac{d}{dt}(x^2 + y^2) = \frac{d}{dt}(2500) \] \[ 2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0 \] Simplifying: \[ x \frac{dx}{dt} + y \frac{dy}{dt} = 0 \] **Step 3: Plug in the Given Values** We have: - \( x = 40 \) m - \( y = 30 \) m - \( \dfrac{dx}{dt} = -1.25 \) m/s Substitute these values into the differentiated equation: \[ (40)\left(-1.25\right) + (30)\frac{dy}{dt} = 0 \] **Step 4: Solve for \( \dfrac{dy}{dt} \)** Compute \( 40 \times (-1.25) \): \[ 40 \times (-1.25) = -50 \] Now, the equation becomes: \[ -50 + 30 \frac{dy}{dt} = 0 \] Add 50 to both sides: \[ 30 \frac{dy}{dt} = 50 \] Divide both sides by 30: \[ \frac{dy}{dt} = \frac{50}{30} \] \[ \frac{dy}{dt} = \frac{5}{3} \text{ m/s} \] **Answer:** \(\displaystyle \frac{dy}{dt} = \frac{5}{3}\) meters per second at that moment. |
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| GPT-4o | To solve for \(\frac{dy}{dt}\) given the information, we can use the relationship between the coordinates and their rates of change, derived from the geometry of the situation. Here are the steps: 1. **Understand the geometry of the problem**: - Julian is jogging around a circular track of radius 50 meters. - His coordinates at the given moment are \((40, 30)\). - The equation of the circle is \(x^2 + y^2 = 50^2\). 2. **Implicit Differentiation**: - Differentiate the equation of the circle with respect to time \(t\): \[ \frac{d}{dt}(x^2 + y^2) = \frac{d}{dt}(50^2) \] \[ 2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0 \] 3. **Solve for \(\frac{dy}{dt}\)**: - Given \(x = 40\), \(y = 30\), and \(\frac{dx}{dt} = -1.25\) m/s, substitute these values into the differentiated equation: \[ 2(40)(-1.25) + 2(30) \frac{dy}{dt} = 0 \] \[ -100 + 60 \frac{dy}{dt} = 0 \] \[ 60 \frac{dy}{dt} = 100 \] \[ \frac{dy}{dt} = \frac{100}{60} = \frac{5}{3} \approx 1.67 \text{ m/s} \] So, \(\frac{dy}{dt}\) at the moment when Julian's coordinates are \((40, 30)\) and his \(x\)-coordinate is changing at a rate of \(-1.25\) m/s is approximately \(1.67\) m/s. |
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| llama3.1-70B-Instruct | A classic related rates problem! Let's break it down step by step: 1. We are given that Julian's x-coordinate is changing at a rate of -1.25 m/s, which means dx/dt = -1.25 m/s. 2. We are also given that Julian's coordinates are (40, 30) at this moment. 3. Since Julian is jogging around a circular track, we can use the equation of a circle: x^2 + y^2 = r^2, where r is the radius of the track (50 m). 4. We can differentiate both sides of the equation with respect to time t, using the chain rule: d/dt (x^2 + y^2) = d/dt (r^2) This gives us: 2x(dx/dt) + 2y(dy/dt) = 0 5. Now, we can plug in the values we know: 2(40)(-1.25) + 2(30)(dy/dt) = 0 Simplifying, we get: -100 + 60(dy/dt) = 0 6. Solving for dy/dt, we get: dy/dt = 100/60 = 5/3 m/s So, at this moment, Julian's y-coordinate is changing at a rate of 5/3 m/s. |
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| llama3.1-405B-Instruct | To find dy/dt, we can use the chain rule and the fact that the derivative of the x-coordinate with respect to time is -1.25 m/s. Let's denote the x-coordinate of Julian's position as x(t) and the y-coordinate as y(t). Since Julian is jogging around a circular track, we know that the distance from the origin to Julian's position is constant and equal to the radius of the track, which is 50 m. Using the Pythagorean theorem, we can write: x(t)^2 + y(t)^2 = 50^2 Differentiating both sides of this equation with respect to time t, we get: 2x(t) \* dx/dt + 2y(t) \* dy/dt = 0 Now, we are given that dx/dt = -1.25 m/s and x(t) = 40 m, y(t) = 30 m. Plugging these values into the equation above, we get: 2(40) \* (-1.25) + 2(30) \* dy/dt = 0 Simplifying and solving for dy/dt, we get: dy/dt = 1.67 m/s Therefore, the rate of change of Julian's y-coordinate with respect to time is 1.67 m/s at the moment when his coordinates are (40, 30). |
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| DeepSeek-Coder-V2 | To solve the problem, we need to find the rate of change of Julian's y-coordinate, dy/dt, given the rate of change of his x-coordinate, dx/dt, and his current coordinates. 1. **Identify the given information:** - Radius of the circular track: \( r = 50 \) m - Julian's coordinates: \( (x, y) = (40, 30) \) - Rate of change of x-coordinate: \( \frac{dx}{dt} = -1.25 \) m/s 2. **Use the equation of the circle:** The equation of the circle with radius 50 centered at the origin is: \[ x^2 + y^2 = 50^2 \] Substituting the given coordinates: \[ 40^2 + 30^2 = 50^2 \] \[ 1600 + 900 = 2500 \] This confirms that the coordinates are on the circle. 3. **Differentiate the equation of the circle with respect to time:** \[ \frac{d}{dt}(x^2 + y^2) = \frac{d}{dt}(50^2) \] \[ 2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0 \] 4. **Substitute the known values:** - \( x = 40 \) - \( y = 30 \) - \( \frac{dx}{dt} = -1.25 \) Substitute these into the differentiated equation: \[ 2(40)(-1.25) + 2(30)\frac{dy}{dt} = 0 \] \[ -100 + 60\frac{dy}{dt} = 0 \] 5. **Solve for \( \frac{dy}{dt} \):** \[ 60\frac{dy}{dt} = 100 \] \[ \frac{dy}{dt} = \frac{100}{60} = \frac{5}{3} \approx 1.67 \text{ m/s} \] Thus, the rate of change of Julian's y-coordinate at this moment is \( \frac{dy}{dt} = \frac{5}{3} \) m/s. |
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| Qwen2-72B-Instruct | To solve this problem, we need to find the rate of change of Julian's y-coordinate (dy/dt) when his x-coordinate is changing at a rate of -1.25 m/s at the point (40, 30). First, we need to find the equation of the circular track. The equation of a circle with radius r and center at the origin (0, 0) is given by: x^2 + y^2 = r^2 Substituting the given radius of 50 m, we get: x^2 + y^2 = 2500 We are given that Julian's x-coordinate is changing at a rate of -1.25 m/s at the point (40, 30). We can find the rate of change of Julian's y-coordinate (dy/dt) by using implicit differentiation. Differentiating both sides of the equation with respect to time t, we get: 2x(dx/dt) + 2y(dy/dt) = 0 Substituting the given values of x, y, and dx/dt, we get: 2(40)(-1.25) + 2(30)(dy/dt) = 0 Simplifying, we get: -100 + 60(dy/dt) = 0 Solving for dy/dt, we get: dy/dt = 100/60 = 5/3 Therefore, the rate of change of Julian's y-coordinate (dy/dt) at the point (40, 30) when his x-coordinate is changing at a rate of -1.25 m/s is 5/3 m/s or approximately 1.67 m/s. |
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| baichuan4 | To solve this problem, we can use the fact that Julian is moving in a circular path and the relationship between the x and y coordinates as he moves along the circle. Since the path is circular with a radius of 50 m, Julian's position at any time can be described by the parametric equations: x(t) = 50 * cos(θ(t)) y(t) = 50 * sin(θ(t)) where θ(t) is the angular position at time t. We are given that the x-coordinate is changing at a rate of -1.25 m/s when Julian's coordinates are (40, 30). This means that the derivative of x(t) with respect to time, evaluated at the specific moment, is -1.25 m/s: dx/dt = -1.25 m/s Now, we need to find dy/dt at the same moment. Taking the derivatives of the parametric equations with respect to time, we get: dx/dt = -50 * sin(θ(t)) * dθ/dt dy/dt = 50 * cos(θ(t)) * dθ/dt We can solve for dθ/dt by using the given value of dx/dt and the x-coordinate at the moment in question: -1.25 = -50 * sin(θ(t)) * dθ/dt dθ/dt = 1.25 / (50 * sin(θ(t))) Now, we need to find θ(t) when Julian's coordinates are (40, 30). We can use the x-coordinate equation to solve for θ(t): 40 = 50 * cos(θ(t)) cos(θ(t)) = 40 / 50 cos(θ(t)) = 0.8 θ(t) = arccos(0.8) Now, we can plug this value of θ(t) into the equation for dy/dt to find the rate of change of the y-coordinate at the moment in question: dy/dt = 50 * cos(arccos(0.8)) * dθ/dt dy/dt = 50 * 0.8 * (1.25 / (50 * sin(arccos(0.8)))) dy/dt = 40 * (1.25 / (50 * sin(arccos(0.8)))) We know that sin(arccos(x)) = √(1 - x^2), so: dy/dt = 40 * (1.25 / (50 * √(1 - 0.8^2))) dy/dt = 40 * (1.25 / (50 * √(1 - 0.64))) dy/dt = 40 * (1.25 / (50 * √(0.36))) dy/dt = 40 * (1.25 / (50 * 0.6)) dy/dt = 40 * (1.25 / 30) dy/dt ≈ 1.67 m/s Therefore, the rate of change of the y-coordinate at the moment when Julian's coordinates are (40, 30) is approximately 1.67 m/s. |
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| doubao | The equation of a circle is x^2 + y^2 = r^2. In this case, r = 50. So x^2 + y^2 = 50^2 = 2500. Differentiating both sides with respect to time t gives: 2x(dx/dt) + 2y(dy/dt) = 0. We are given that dx/dt = -1.25 and x = 40, y = 30. Substituting these values into the equation: 2*40*(-1.25) + 2*30*(dy/dt) = 0. -100 + 60*(dy/dt) = 0. 60*(dy/dt) = 100. dy/dt = 100/60 = 5/3 m/s. |
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| minimax | To find the rate at which Julian's y-coordinate is changing (dy/dt) when his x-coordinate is changing at a rate of -1.25 m/s at the point (40, 30), we can use the relationship between the x-coordinate and the y-coordinate on the circular track. Since Julian is jogging on a circular track, the relationship between his x and y coordinates at any point in time can be described by the equation of a circle centered at the origin: \[ x^2 + y^2 = r^2 \] where \( r \) is the radius of the track. Given that the radius \( r \) is 50 m, the equation becomes: \[ x^2 + y^2 = 50^2 \] Differentiating both sides of this equation with respect to time \( t \) gives us: \[ 2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0 \] We are given that \( \frac{dx}{dt} = -1.25 \) m/s when \( x = 40 \) m and \( y = 30 \) m. We can plug these values into the differentiated equation to solve for \( \frac{dy}{dt} \): \[ 2(40)(-1.25) + 2(30) \frac{dy}{dt} = 0 \] \[ -100 + 60 \frac{dy}{dt} = 0 \] Now, solve for \( \frac{dy}{dt} \): \[ 60 \frac{dy}{dt} = 100 \] \[ \frac{dy}{dt} = \frac{100}{60} \] \[ \frac{dy}{dt} = \frac{5}{3} \] \[ \frac{dy}{dt} = 1.666... \] So, Julian's y-coordinate is changing at a rate of approximately 1.67 m/s when his x-coordinate is changing at a rate of -1.25 m/s at the point (40, 30). |
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| kimi | To solve this problem, we can use the relationship between the coordinates on a circle and the derivatives of those coordinates with respect to time. Julian is jogging around a circular track, so at any given point, his position can be described by the parametric equations of a circle: \[ x = r \cos(\theta) \] \[ y = r \sin(\theta) \] where \( r \) is the radius of the circle (50 m in this case) and \( \theta \) is the angle from the positive x-axis. Given that Julian's x-coordinate is changing at a rate of -1.25 m/s, we have: \[ \frac{dx}{dt} = -1.25 \, \text{m/s} \] We also know his coordinates at a particular moment are (40, 30), which means: \[ x = 40 \] \[ y = 30 \] Using the parametric equations, we can find \( \theta \) at this moment: \[ 40 = 50 \cos(\theta) \] \[ 30 = 50 \sin(\theta) \] Dividing the second equation by the first gives us: \[ \frac{30}{40} = \frac{50 \sin(\theta)}{50 \cos(\theta)} \] \[ \frac{3}{4} = \tan(\theta) \] Now, we can find \( \theta \) by taking the arctangent: \[ \theta = \arctan\left(\frac{3}{4}\right) \] Next, we can find the derivatives of x and y with respect to time using the chain rule: \[ \frac{dx}{dt} = -r \sin(\theta) \frac{d\theta}{dt} \] \[ \frac{dy}{dt} = r \cos(\theta) \frac{d\theta}{dt} \] We already have \( \frac{dx}{dt} \) and we can express \( \frac{dy}{dt} \) in terms of \( \frac{dx}{dt} \) and \( \theta \): \[ \frac{dy}{dt} = -\frac{y}{x} \frac{dx}{dt} \] Plugging in the values we have: \[ \frac{dy}{dt} = -\frac{30}{40} (-1.25) \] \[ \frac{dy}{dt} = -\frac{3}{4} (1.25) \] \[ \frac{dy}{dt} = -0.75 \times 1.25 \] \[ \frac{dy}{dt} = -0.9375 \, \text{m/s} \] So, \( \frac{dy}{dt} \) at this moment is \( -0.9375 \, \text{m/s} \). |
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| wenxinyiyan | To solve this problem, we will use the concept of differentiation and the Pythagorean theorem to find the relationship between the x and y coordinates of Julian's position on the circular track. 1. **Given Information**: * Radius of the track (r) = 50 m * Rate of change of x-coordinate (dx/dt) = -1.25 m/s * Current coordinates of Julian = (40, 30) 2. **Relationship between x and y**: * Since Julian is jogging around a circular track, his position (x, y) at any time will satisfy the equation of a circle: x^2 + y^2 = r^2. * Given r = 50, the equation becomes x^2 + y^2 = 2500. 3. **Differentiation**: * Differentiating the equation x^2 + y^2 = 2500 with respect to time (t), we get: + 2x(dx/dt) + 2y(dy/dt) = 0 * This equation relates the rates of change of x and y with respect to time. 4. **Solving for dy/dt**: * Rearranging the differentiated equation, we get: + dy/dt = - (x/y) * (dx/dt) * At the point (40, 30), x = 40 and y = 30. Also, dx/dt = -1.25 m/s. * Substituting these values into the equation for dy/dt: + dy/dt = - (40/30) * (-1.25) + dy/dt = 5/3 * 1.25 + dy/dt = 2.0833 m/s (approx.) Therefore, at the moment when Julian's coordinates are (40, 30), the rate of change of his y-coordinate (dy/dt) is approximately 2.0833 m/s. |
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| zhipu | To solve this problem, we need to use related rates and the equation of a circle. Julian is jogging around a circular track with a radius of 50 meters, so the equation of the circle representing the track is: \[ x^2 + y^2 = 50^2 \] Given Julian's coordinates \((x, y) = (40, 30)\), we can verify that these points lie on the circle by plugging them into the equation: \[ 40^2 + 30^2 = 1600 + 900 = 2500 \] \[ 50^2 = 2500 \] Since both sides are equal, the coordinates do indeed lie on the circle. We are given that the x-coordinate is changing at a rate of \(-1.25 \, \text{m/s}\), which is \(\frac{dx}{dt}\). We need to find \(\frac{dy}{dt}\), the rate at which the y-coordinate is changing. To find \(\frac{dy}{dt}\), we differentiate the circle equation with respect to time \(t\): \[ \frac{d}{dt}(x^2 + y^2) = \frac{d}{dt}(50^2) \] Since the radius of the circle is constant, the right side of the equation is zero: \[ 2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0 \] We can divide through by 2 to simplify: \[ x \frac{dx}{dt} + y \frac{dy}{dt} = 0 \] Now we solve for \(\frac{dy}{dt}\): \[ y \frac{dy}{dt} = -x \frac{dx}{dt} \] \[ \frac{dy}{dt} = -\frac{x}{y} \frac{dx}{dt} \] We substitute the given values \(x = 40\), \(y = 30\), and \(\frac{dx}{dt} = -1.25\): \[ \frac{dy}{dt} = -\frac{40}{30} (-1.25) \] \[ \frac{dy}{dt} = -\frac{4}{3} (-1.25) \] \[ \frac{dy}{dt} = \frac{5}{3} \] So, the rate at which Julian's y-coordinate is changing at the moment his coordinates are (40, 30) is: \[ \frac{dy}{dt} = \frac{5}{3} \, \text{m/s} \] This means that Julian's y-coordinate is increasing at a rate of \( \frac{5}{3} \) meters per second at that instant. |
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| gpto1 | **Solution:** We are given: - Julian is jogging around a circular track of radius \( r = 50 \) meters. - His position at a certain moment is \( (x, y) = (40, 30) \). - At that moment, his \( x \)-coordinate is changing at a rate of \( \dfrac{dx}{dt} = -1.25 \) m/s. - We are asked to find \( \dfrac{dy}{dt} \) at that moment. **Step 1: Write the Equation of the Circle** Since Julian is jogging around a circular track of radius 50 m centered at the origin, the equation of the circle is: \[ x^2 + y^2 = r^2 \] \[ x^2 + y^2 = 50^2 \] \[ x^2 + y^2 = 2500 \] **Step 2: Differentiate Both Sides with Respect to Time \( t \)** We will use implicit differentiation with respect to time \( t \): \[ \frac{d}{dt}(x^2 + y^2) = \frac{d}{dt}(2500) \] \[ 2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0 \] Simplifying: \[ x \frac{dx}{dt} + y \frac{dy}{dt} = 0 \] **Step 3: Plug in the Given Values** We have: - \( x = 40 \) m - \( y = 30 \) m - \( \dfrac{dx}{dt} = -1.25 \) m/s Substitute these values into the differentiated equation: \[ (40)\left(-1.25\right) + (30)\frac{dy}{dt} = 0 \] **Step 4: Solve for \( \dfrac{dy}{dt} \)** Compute \( 40 \times (-1.25) \): \[ 40 \times (-1.25) = -50 \] Now, the equation becomes: \[ -50 + 30 \frac{dy}{dt} = 0 \] Add 50 to both sides: \[ 30 \frac{dy}{dt} = 50 \] Divide both sides by 30: \[ \frac{dy}{dt} = \frac{50}{30} \] \[ \frac{dy}{dt} = \frac{5}{3} \text{ m/s} \] **Answer:** \(\displaystyle \frac{dy}{dt} = \frac{5}{3}\) meters per second at that moment. |
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| deepseekv3 | To find the rate at which Julian's \( y \)-coordinate is changing (\( \frac{dy}{dt} \)) at the given moment, we'll use the relationship between his position on the circular track and the rates of change of his coordinates. ### Step 1: Understand the Problem Julian is jogging around a circular track with a radius of 50 meters. At a specific moment, his coordinates are \( (40, 30) \), and his \( x \)-coordinate is changing at a rate of \( -1.25 \, \text{m/s} \) (\( \frac{dx}{dt} = -1.25 \, \text{m/s} \)). We need to find \( \frac{dy}{dt} \) at this moment. ### Step 2: Use the Equation of the Circle The equation of the circle with radius \( r \) centered at the origin is: \[ x^2 + y^2 = r^2 \] Given \( r = 50 \), the equation becomes: \[ x^2 + y^2 = 2500 \] ### Step 3: Differentiate Both Sides with Respect to Time To find the relationship between \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \), differentiate both sides of the equation with respect to time \( t \): \[ \frac{d}{dt}(x^2) + \frac{d}{dt}(y^2) = \frac{d}{dt}(2500) \] Using the chain rule: \[ 2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0 \] ### Step 4: Solve for \( \frac{dy}{dt} \) Rearrange the equation to solve for \( \frac{dy}{dt} \): \[ 2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0 \\ 2y \frac{dy}{dt} = -2x \frac{dx}{dt} \\ \frac{dy}{dt} = -\frac{x}{y} \frac{dx}{dt} \] ### Step 5: Plug in the Given Values At the moment in question, Julian's coordinates are \( (40, 30) \) and \( \frac{dx}{dt} = -1.25 \, \text{m/s} \). Substitute these values into the equation: \[ \frac{dy}{dt} = -\frac{40}{30} \times (-1.25) \\ \frac{dy}{dt} = \frac{40}{30} \times 1.25 \\ \frac{dy}{dt} = \frac{4}{3} \times 1.25 \\ \frac{dy}{dt} = \frac{5}{3} \\ \frac{dy}{dt} \approx 1.67 \, \text{m/s} \] ### Final Answer At the moment when Julian's coordinates are \( (40, 30) \) and \( \frac{dx}{dt} = -1.25 \, \text{m/s} \), the rate of change of his \( y \)-coordinate is: \[ \boxed{\dfrac{5}{3} \, \text{m/s}} \] |
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